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3 years ago

Please help! Not even sure how to start with this one or how to calculate

Mathematics

1 answer:

vivado [14]3 years ago

4 0

Remember when you learning about the "Least Common
Multiple" and you wondered when in the world you would
ever need to use it ?
Well, here you are ! Hello !

The next time they will cross the starting line together is the
LCM (Least Common Multiple) of 30 and 40 .

Do you remember how to find it ?
Here's one easy way:

Count by 30s until you come to a number that's divisible by 40 .

Laura's first lap. . . . . . . . 30 seconds . . . not divisible by 40
Laura's two laps . . . . . . . 60 seconds . . . not divisible by 40
Laura's three laps. . . . . . 90 seconds . . . not divisible by 40
Laura's four laps. . . . . . 120 seconds . . . divisible by 40

After 120 seconds, Laura has completed four laps,
Shelby has completed only three laps.

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Read 2 more answers

Change the subject of the formula L = v 4kt - p to k.

Romashka [77]

Answer:

$\boxed{k = \frac{L^2 + p}{4t}}$

General Formulas and Concepts:

Algebra I

Basic Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Step-by-step explanation:

Step 1: Define

Identify given.

$\displaystyle L = \sqrt{4kt - p}$

Step 2: Solve for k
We can use equality properties to help us rewrite the equation to get k as our subject:

Let's first square both sides:

$\displaystyle\begin{aligned}L = \sqrt{4kt - p} & \rightarrow L^2 = \big( \sqrt{4kt - p} \big) ^2 \\& \rightarrow L^2 = 4kt - p \\\end{aligned}$

Next, add p to both sides:

$\displaystyle\begin{aligned}L = \sqrt{4kt - p} & \rightarrow L^2 = \big( \sqrt{4kt - p} \big) ^2 \\& \rightarrow L^2 = 4kt - p \\& \rightarrow L^2 + p = 4kt \\\end{aligned}$

Next, divide 4t by both sides:

$\displaystyle\begin{aligned}L = \sqrt{4kt - p} & \rightarrow L^2 = \big( \sqrt{4kt - p} \big) ^2 \\& \rightarrow L^2 = 4kt - p \\& \rightarrow L^2 + p = 4kt \\& \rightarrow \frac{L^2 + p}{4t} = k \\\end{aligned}$